Separation Process Principles- 2n - Seader & Henley - Solutions Manual

13 6 the mole fraction of benzene in the cumulative

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: : Time for the mole fraction of A in the instantaneous distillate to fall to 0.40. Analysis: Base the calculations on A, which is the more volatile component. Initially, the distillate mole fraction in equilibrium with the initial charge is obtained from Eq. (4-8) for constant relative volatility, α: y= αx 2.5(0.30) = = 0.5172 1 + x (α − 1) 1 + 0.30(2.5 − 1) (1) When the mole fraction of A in the instantaneous distillate = yD = 0.40, the mole fraction of A in the liquid in the still is obtained form a rearrangement of Eq. (1), x= 0.40 y = = 0.2105 α + y (1 − α ) 2.5 + 0.40(1 − 2.5) Modify Eq. (13-1) to include the constant feed added to the still, noting that, with the above assumptions, W = total moles in the still = constant = W0 = 20 lbmole, dW/dt = 0, and the distillate rate, call it D = molar feed rate, F: Fx F − W dx = Dy D = Fy D dt Rearranging, dx F x F − y D = dt W0 (2) Eq. (2) is integrated as follows, noting that yD = y. Substituting Eq. (1) into Eq. (2) gives, F 10 dt = dt = W0 20 xF − t 0.2105 0 0.305 0.5 dt = 0.50t = dx dx (1 + 15x )dx . = = 2.5x αx 0.30 − 2.05x 0.30 − 1 + x (α − 1) 1 + 15x . 1 + 15x )dx . = 0.585 0.305 − 2.05x Therefore, t = 0.585/0.5 = 1.17 h Exercise 13.9 Subject: Batch distillation of a mixture of isopropyl alcohol (P) and water (W) in a column with 2 equilibrium stages and a reboiler, under conditions of a constant reflux ratio. Given: Feed contains 40 mol% P and 60 mol% W. Distillation at 1 atm with a reflux of L/V = 0.9. Vapor-liquid equilibrium data in Exercise 13.2. Assumptions: Perfect mixing in the still. No holdup on the stages or in the condenser. Find: Compositions of the residue in the still and cumulative distillate when 70 mol% of the charge has been distilled (W/W0 = 0.30). Analysis: Make calculations in terms of P, the more volatile component. Eq. (13-2) applies, where yD is the mole fraction of P in the vapor leaving the top stage, and xW is the mole fraction of P in the liquid in the reboiler. ln W0 1 = ln = 1.204 = W 0.3 0.40 xW dxW y D − xW (1) The relationship between yD and xW is obtained from a McCabe-Thiele diagram by drawing a series of operating lines of slope = L/V = 0.9. For each operating line, starting from the intersection with the 45o line, which is yD, 3 stages are stepped off to determine the corresponding xW. A typical construction that starts from yD = 0.65 is shown on the next page, where the xW = 0.37. Other sets of values are given in the following table, which also includes values of the integrand, f = 1/(yD - xW), from which the integral is evaluated by the trapezoidal rule with variable increments of ∆xW. For example, the increment of the integral from x1 to x2 is (x1 - x2)(f1/2 + f2/2). The increments are summed over the region from xW = 0.40 to the value corresponding to 70 mol% distilled. xW yD from McCabe-Thiele f = 1/(yD - xW) 0.400 0.370 0.260 0.170 0.080 0.015 0.008 0.002 0.66 0.65 0.645 0.64 0.63 0.60 0.55 0.50 3.90 3.61 2.60 2.13 1.82 1.71 1.85 2.01 Increment of Int...
View Full Document

This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

Ask a homework question - tutors are online